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Theorem 1.2. If X ⊆ Rd is infinite with |X| regular, and 2 ≤ a ≤ d+1, then there is an a-rainbow Y ⊆ X with |Y | = |X|.

Erd˝os also gives the following example:

Theorem 1.3. If λ ≤ 2ℵ0 is singular then there is X ⊆ Rd of size λ, such that there is no 3-rainbow Y ⊆ X with |Y | = λ.

2 Proof. Write cof(κ) = λ. Let (ℓα : α < λ) be λ-many parallel lines in R . Let (κα : α < λ) be a cofinal sequence of regular cardinals in κ. Choose Xα ⊆ ℓα of cardinality κα and let X = Xα. Sα<λ Let Y ⊆ X have cardinality κ. We claim that Y cannot be 3-rainbow. Indeed, write Yα = Y ∩Xα = Y ∩ ℓα. Then there must be cofinally many α < λ with Yα infinite, as otherwise |Y |≤ κα + λ < λ for some α < λ. Thus we can find α < β < λ such that Yα and Yβ are both infinite. Let v0, v1 be two distinct points in Yα, and let w0, w1 be two distinct points in Yβ. Then the triangles (v0, v1, w0) and (v0, v1, w1) have the same nonzero area.

We are interested in strengthenings and generalizations of Theorem 1.2 for uncountable sets. We will give stronger canonization results than just a-rainbow. Namely, say that X is strongly a-rainbow if all a-subsets of X yield distinct, nonzero volumes, and say that X is strictly a-rainbow if X is strongly a′-rainbow for all a′ ≤ a, and X is a subset of an a − 1-dimensional hyperplane. (In particular, all a + 1-subsets of X have volume 0.) As an example, if an,i : n<ω,i

2 results of Section 2. We prove in Theorem 3.3 that for every regular cardinal κ ≤ 2ℵ0 , and for every X ⊆ Rd of size κ, there is some X′ ⊆ X of size κ and some 2 ≤ a ≤ d + 1 such that X is strictly a-rainbow. We proceed as follows: given X ⊆ Rd of size κ, we obtain a sufficiently indiscernible Y ⊆ X using Theorem 2.4, using the stability of (C, +, ·, 0, 1). Then we apply geometric arguments to argue that Y is strictly a-rainbow for some a. We note that it is possible to prove Theorem 3.3 directly, similarly to Theorem 1.2. For singular cardinals, we know from Theorem 2.6 that there is some finite list of possible con- figurations, although we cannot identify it explictly. We are at least able to give some information about what the configurations look like in Theorem 3.4; in particular, they are all 2-rainbow, and so we recover Erd˝os’s Theorem 1.1. In Section 4, we consider what happens for X ⊆ Rd which is reasonably definable. Our main result is Theorem 4.3: if P ⊆ Rd is perfect, then there is a perfect Q ⊆ P and some a ≤ d +1 such that Q is strictly a-rainbow. Our main tool is a Ramsey-theoretic result of Blass [1] concerning colorings of perfect trees. In Section 5, we show it is independent of ZF whether or not every uncountable subset of R has an uncountable 2-rainbow subset. In this paper we work in ZFC, with the exception of Section 4, which is in ZF + DC.

2 Some remarks on indiscernibles

We first review the notion of local indiscernibility, following Shelah [10]. T will always be a complete first order theory in a countable language. Suppose ∆ is a collection of formulas of T , M |= T and A ⊆ M. Given a finite tuple b from <ω M, define tp∆(b/A) to be the set of all formulas φ(x, a) such that a ∈ A and φ(x, y) ∈ ∆ and M |= φ(b, a). d Suppose also that I is an index set, and (ai : i ∈ I) is a sequence from M for some d<ω. Then:

• We say that (ai : i ∈ I) is ∆-indiscernible over A if: given i0,...,in−1 all distinct ele-

ments of I, and given j0,...,jn−1 also distinct elements from I, then tp∆(ai0 ,..., ain−1 /A)=

tp∆(bi0 ,..., bin−1 ). In this case the indexing doesn’t matter and so we also say that {ai : i ∈ I} is indiscernible over A.

• If I is linearly ordered, then we say that (ai : i ∈ I) is ∆-order-indiscernible over A if: for

every i0 <...

When we do not mention A, we mean A = ∅. The following is an easy application of Ramsey’s theorem (as recorded for instance in Lemma 2.3 of Chapter I of [10]:

Theorem 2.1. Suppose T is a complete first order theory in a countable language, and ∆ is a finite collection of formulas of T . Suppose M |= T , and (an : n<ω) is an infinite sequence from d M . Then there is some infinite subsequence (an : n ∈ I) which is ∆-order indiscernible.

3 We will be mainly interested in the case where T is stable. In this case, the following is part of Theorem 2.13 of Chapter II of [10]:

Theorem 2.2. Suppose T is a stable complete first order theory in a countable language, and ∆ is a finite collection of formulas of T . Suppose M |= T , and (an : n<ω) is an infinite sequence d from M . Then (an : n<ω) is order-indiscernible if and only if it is indiscernible (in fact this characterizes stability). Hence, if X ⊆ M d is infinite, then there is an infinite, ∆-indiscernible Y ⊆ X.

We are interested in generalizations of Theorem 2.2 to the case where X has uncountable cardinality κ. Shelah has proved several results along these lines for regular cardinals; we give two versions. The first requires T to be ω-stable, and gets full indiscernibility. See Remark 2 after Theorem 2.8 from Chapter 1 of [10].

Theorem 2.3. Let T be an ω-stable theory (we can suppose in a countable language). Let κ be a regular uncountable cardinal and let d<ω. Then whenever M |= T , A ⊆ M has size less than κ, and X ⊆ M d has size ≥ κ, we can find some finite sequence a ∈ M, and we can find some stationary type p(x) ∈ Sd(a), such that there is some Y ⊆ X of size κ which is a set of independent realizations of p(x)|Aa. In particular, Y is indiscernible over A.

Proof. For the reader’s convenience we provide a proof. We can suppose T = T eq, and thus that we can code finite tuples as single elements. Also we can suppose that |X| = κ. Enumerate X = (aα : α < κ). For each α < κ, write Xα = acl (A ∪ {aβ : β < α}), and choose a formula φα(x) over Xα of the same Morley rank as tp(aα/Xα), and of Morley degree 1. By Fodor’s lemma, we can find S ⊆ κ stationary such that φα(x) =

φβ(x)= φ(x) for all α, β ∈ S. Choose α∗ large enough that φ(x) is over Xα∗ . Write φ(x)= φ(x, a) for some a ∈ Xα. Let p(x) be the unique type over a containing φ(x, a) and of the same Morley rank; then for all α∗ ≤ α ∈ S, tp(aα/Xα) is the unique non-forking extension of p(x) to Xα. From this it follows easily that Y := {aα : α ∈ S\α∗} is as desired.

The second version applies to any stable theory, but only gives local indiscernibility. It is Theorem 2.19 of Chapter II of [10], and it strictly generalizes the final claim of Theorem 2.2.

Theorem 2.4. Let T be a stable theory and let ∆ be a finite set of formulas. Let κ be a regular cardinal and let d<ω. Then whenever M |= T , A ⊆ M has size less than κ, and X ⊆ M d has size ≥ κ, then there is some Y ⊆ X of size κ such that Y is ∆-indiscernible over A.

When κ is singular, the na¨ıve generalization of Theorem 2.3 fails, by Theorem 1.3. (In fact, one can easily modify this example to work in the theory of equality.) The problem here is the presence of an equivalence relation on X that has fewer than κ classes, with each class of size less than κ, and the behavior of elements in distinct classes differs from the behavior of elements in the same class. In fact, we show this is the only obstruction. We wish to formalize the notion of “indiscernible up to an equivalence relation.” This a special case of generalized indiscernibles, introduced by Shelah in [10] Section VII.2, and further analyzed (with slightly varying definitions) in several subsequent papers, e.g. [5]. (In these works, the focus

4 is on using these generalized indiscernibles to build Ehrenfeucht-Mostowski models; our interest is different, in that we want to extract generalized indiscernibles from a given X.) Suppose T is a complete first order theory, and ∆ is a collection of formulas. Suppose M |= T , and A ⊆ M, and X ⊆ M d for some d. Finally suppose E is an equivalence relation on X. Then X is (∆, E)-indiscernible over A if for every a0,..., an−1, b0,..., bn−1 sequences from X with each ai 6= aj and each bi 6= bj, if for every i

Theorem 2.5. Let T be an ω-stable theory, and let κ be a singular cardinal of cofinality λ > ℵ0, and let d<ω. Then whenever M |= T , A ⊆ M has size less than cof(κ), and X ⊆ M d has size ≥ κ, there is some Y ⊆ X of size κ and some equivalence relation E on Y , such that E has at most λ-many equivalence classes, with each equivalence class infinite, and such that Y is E-indiscernible over A.

Proof. We can suppose T = T eq, and thus that d = 1. Write X as the of Xα : α < λ, where each |Xα| = κα < κ is a successor cardinal bigger than |A|, and κα < κβ whenever α < β. By applying Theorem 2.3 to each Xα and then 1 pruning, we can suppose there is some aα ∈ M and some stationary p(x) ∈ S (aα), such that Yα is an independent set of realizations of p(x)| A ∪ Xβ . Define the equivalence relation E on  Sβ<α  X by: aEb iff a, b are in the same Xα. For each α < λ, choose φα(x, a) ∈ pα(x) of the same Morley rank as p(x), and of Morley degree 1. By further pruning, we can suppose φα(x,y)= φβ(x,y) for all α, β < λ. 1 Apply Theorem 2.3 and prune to get some a∗ and some stationary type q(x) ∈ S (a∗), such that {aα : α < λ} is an independent set of realizations of q(x)|Aa∗. We claim now that X is E-indiscernible. To check this, it is convenient to discard all but α countably many elements of each Xα. Thus enumerate each Xα = {bn : n<ω}. Now each α (b : n<ω) is a Morley sequence in pα(x)| A, aα, Xβ , and (aα : α < λ) is a Morley sequence n  Sβ<α  in q(x)|Aa∗. It follows by typical nonforking arguments that for each α, Xα is indiscernible over α σ(α) A ∪ Xβ; also, for each permutation σ of λ, the permutation σ∗ of X defined by σ∗(b )= bn Sβ6=α n is partial elementary over A. From these two facts it is easy to check that X is E-indiscernible.

Before the following theorem, we need some notation. Given X a set and r<ω, X denotes r  the r-element subsets of X. When X is a set of ordinals, each s ∈ X has a canonical increasing r  enumeration; thus we can identify X ⊆ Xr. r  Theorem 2.6. Let T be a stable theory, and let ∆ be a finite collection of formulas of T . Let κ be a singular cardinal, and let d<ω. Then whenever M |= T , A ⊆ M has size less than cof(κ),

5 and X ⊆ M d has size ≥ κ, there is some Y ⊆ X of size κ and some equivalence relation E on Y , such that E has ℵ0-many classes, with each equivalence class infinite, and such that Y is (∆, E)-indiscernible over A.

Proof. Again, we can suppose T = T eq and d = 1. Let r be the maximum of the arities of formulas of ∆. Write λ = cof(κ) and write X as the disjoint union of Xα : α < λ, where each |Xα| = κα < κ ′ is a successor cardinal bigger than |A|, and κα < κα′ whenever k < k . i Then by many applications of Theorem 2.3, we can find distinct aα,j : α < λ, i, j < r such that i i each aα,j ∈ Xα, and we can find Yα : α<ω,i ≤ r, such that:

i+1 i • Each Yα ⊆ Yα ⊆ Xα;

i • Each Yα has size κk;

i i i+1 • Each aα,j ∈ Yα\Yα ;

′ • Each Y i is indiscernible over A ∪ Y i ∪ {ai : β < λ, i′ < i, j < r}. α Sβ<α β β,j i For each α < λ, let bα = (aα,j : i, j < r). By applying Theorem 2.4, we can suppose that (bα : α < λ) is ∆-indiscernible over A. Given an injective s : r → λ, let ps(xi,j : i, j < r) = r−1−i tp∆(as(i),j : i, j < r). This is a ∆-type of a finite tuple over the ; in particular it is equivalent to a formula, but it is more convenient to write it as a type. Note that by ∆- ′ indiscernibility, for all s,s , ps(xi,j : i, j < r) = ps′ (xi,j : i, j < r), so we can drop the subscripts and refer to just p(xi,j : i, j < r). r For each α, write Yα = Y and write Y = Yα. We claim that Y is (∆, E ↾Y )-indiscernible. α Sα Note that for all s ∈ ω and for all distinct a : i, j < r, with a ∈ Y , we have that r s(i),j s(i),j s(i) M |= p(as(i),j : i, j < r). This is because, starting from (as(r−1),j : j < r) and moving downwards, r−1−i r−1−i we can shift each (as(i),j : j < r) to (as(i),j : j < r); when moving (as(i),j : j < r) to (as(i),j : j < r) r−1−i we are using the indiscernibility hypothesis on Ys(i) . We now need to take care of the fact that we are only looking at the increasing enumeration of s in the above. Choose distinct (aα,j : α < λ, j < r) with each aα,j ∈ Yk. Write aα = (aα,j : j < r). By the preceding, we have that (aα : α < λ) is order-∆-indiscernible; but by Theorem 2.2, this implies that (aα : α < λ) is fully indiscernible. In particular, given any injective sequence s : r → λ, and given distinct as(i),j : i, j < r, with as(i),j ∈ Ys(i), we have that M |= p(as(i),j : i, j < r) (since we could have chosen (aα : α < λ) to cover range(s)). From this it follows easily that Y is (∆, E ↾Y )-indiscernible.

The following theorem is an easy consequence of Theorems 2.4 and 2.6.

Theorem 2.7. Suppose T is stable, M |= T , ∆ is a finite set of formulas, d<ω, and κ is an d infinite cardinal. Then there is a finite list (Ci : i < i∗) such that: each Ci ⊆ M has size κ, and for d every X ⊆ M of size κ, there is some Y ⊆ X of size κ and some i < i∗ such that tp∆(Ci)= tp∆(Y ) (i.e. there is a bijection f : Ci → Y that preserves ∆-formulas). If κ is regular, then each Ci is

6 ∆-indiscernible; otherwise, each Ci is (∆, Ei)-indiscernible for some equivalence relation Ei on Ci with infinitely many classes, each class infinite.

We give a purely finitary analogue of Theorem 2.6. First, given n,r,c < ω, an equivalence relation E on n and a function f : n → c, say that X ⊆ n is E-homogeneous for f if for all r s,t ∈ X , if s(i)Es(j) iff t(i)Et(j) for all i, j < r, then f(s) = f(t). Also, given A ⊆ Y ⊆ n, say r  that A is convex in Y if whenever n0

Theorem 2.8. Suppose K,L,r,c < ω are given. Then there are K∗,L∗ <ω large enough so that whenever n ≥ K∗ ·L∗, and whenever E is an equivalence relation on n with at least K∗ many classes, of size at least L , and whenever f : n → c, there is some X ⊆ n which is E-homogeneous for f ∗ r such that E ↾X has at least K many classes, each convex in X and of size at least L.

First, we want the following claim.

Claim. Suppose K,L < ω are given. Then there are K∗,L∗ < ω large enough so that whenever n ≥ K∗ · L∗, and whenever E is an equivalence relation on n with at least K∗ many classes, each of size at least L∗, there is some X ⊆ n such that E ↾X has at least K many classes, each convex in X and of size at least L.

Proof. Choose K such that K → (K)2. Choose L such that L → (L2)2 . ∗ ∗ 2 ∗ ∗ (2K∗)! Suppose E is given. We can suppose E is an equivalence relation on n with exactly K∗-many classes, each of size exactly L∗; so n = K∗L∗. Let (Xk : k < K∗) list the equivalence classes of E in some order. For each k < K∗, ℓ

7 n r fs : r−|s| → c is constant on X. For the purposes of this theorem, say that n → (m)c,p if:  n whenever f : r → c, and whenever A ⊆ n has size at most p, there is X ⊆ n of size m which is  r ′ pr2r r homogeneous for f over A. Easily, if n → (m)c′ , where c = c , then n → (m)c,p. We now are ready to prove Theorem 2.8.

r ′ r2r Proof. We follow the proof of Theorem 2.5. Choose K∗ so that K∗ → (K + r)c′ , where c = c . i Choose numbers (Lk : k < K∗, −1 ≤ i ≤ r) such that:

i−1 i r ′ i • For all 0 ≤ i ≤ r and for all k < K∗, L → (L ) ′ , where c = i · r · K∗ + ′ L ′ . k k c,c Pk

i+1 i • For each 0 ≤ i < r and k < K∗, Yk ⊆ Yk ⊆ Xk; i i • For each 0 ≤ i ≤ r and k < K∗, |Yk | = Lk;

i i i+1 • For each 0 ≤ i < r, k < K∗ and j < r, ak,j ∈ Yk \Yk ;

i i′ ′ ′ ′ i • Suppose 0 ≤ i ≤ r, k < K∗; set A = ′ Y ′ ∪ {a ′ ′ : 0 ≤ i < i, k < K∗, j < r}. Then Y Sk

I Given a = (ai : i ∈ I) an injective sequence from n, by tp(s) we mean the function r → c {ai:i∈I}  induced from f : r → c. 2r ′ r  K∗ ′ K∗ Write c = c (as in the definition of K∗) and choose g : r → c so that for all s ∈ r , r−1−i ′  g(s) codes tp(as(i),j : i, j < r). By choice of K∗, we can find I ⊆ K∗ of size K + r, which is homogeneous for g. Let I be the first K-many elements of I′. Write Y = Y r. Then we claim Sk∈I k Y is E-homogeneous for f. Indeed, suppose s,t ∈ Y , such that for all i, j < r, s(i)Es(j) iff t(i)Et(j). We want to show r  that f(s)= f(t). Write s as the disjoint union of its equivalence classes listed in increasing order: s = i

8 3 Getting Large Strictly Rainbow Sets

In this section, we are interested in applying the machinery of the previous section to analyze a- ary-volumes of subsets of R. Recall that we are interested in the following kind of problem: given X ⊆ Rd infinite and given a ≤ d + 1, can we find Y ⊆ X with |Y | = |X|, such that all distinct a-element sets from Y give distinct volumes? The most natural structure to work in for this would be (R, +, ·, 0, 1), but the first-order theory of this structure is unstable. Thus we view R ⊆ C and work in the larger field (C, +, ·, 0, 1) instead; it is well known that its first order theory, ACF0, is ω-stable. We could alternatively look under the hood of Theorem 2.4 and note that it applies to Th(R, +, ·, 0, 1) provided ∆ is taken to be a set of quantifier-free formulas, but this really amounts to the same thing. First we need to show that the relevant notions of a-ary volumes are definable in C.

d a+1 Definition 3.1. Let a ≤ d + 1 <ω. Note that for (v0, . . . , va−1) ∈ (R ) , the a − 1-ary volume 1 of (v0, . . . , va−1) is C |qa(v0, . . . , va−1)| for some constant C and some polynomial qa(v0, . . . , va−1) (namely the Cayley-Menger determinant). Thus (v0, . . . , va−1) has a−1-ary volume 0 iff qa(v0, . . . , va−1)= 0. Let pa(v0, . . . , va−1, w0, . . . , wa−1) be the polynomial in 2da variables given by (qa(v0, . . . , va−1) − qa(w0, . . . , wa−1))(qa(v0, . . . , va−1)+ qa(w0, . . . , wa−1)). Note that for (v0, . . . , va−1), (w0, . . . , wa−1) d a from (R ) , we have that their a−1-ary volumes are the same if and only if pa(v0, . . . , va−1, w0, . . . , wa−1)= 0. Let ∆d be the finite collection of formulas of ACF0 of the form: qa(v) = 0, or pa(v, w) = 0, for a ≤ d + 1.

Thus, Theorem 2.7 gives, for each infinite cardinal κ, a finite basis of all possible ∆d-configurations, and we wish to understand the ones that can be embedded in Rd. In the case where κ is regular, we succeed completely with Theorem 3.3. First we need the following lemma.

Lemma 3.2. Let d be given. Suppose X ⊆ Rd is infinite, and when viewed as a subset of Cd, is ∆d-indiscernible. Then X is strictly a-rainbow for some 2 ≤ a ≤ d + 1.

Proof. Let a be largest so that for some or any v0, . . . , va−1 ∈ X, we have that qa(v0, . . . , va−1) 6= 0. Then clearly X is a subset of the a − 1-dimensional hyperplane spanned by any a elements from X, and for every 2 ≤ a′ ≤ a, every a′-subset of X has nondegenerate volume, so it suffices to ′ ′ show that X is a rainbow for all 2 ≤ a ≤ a. Suppose not; say (v0, . . . , va′−1) and (w0, . . . , wa′−1) X are from a′ of the same volume, that is qa′ (v0, . . . , va′ − 1) = ±qa′ (w0, . . . , wa′−1). We can  ′ suppose there is ℓ < a − 1 such that vi = wi for all i<ℓ, and vi 6= wj for any i, j ≥ ℓ. By a′ − ℓ-many applications of indiscernibility of X, we can suppose ℓ = a′ − 2, or in other words: for all v0, . . . , va′−1, va′ from X distinct, qa′ (v0, . . . , va′−2, va′−1) = ±qa′ (v0, . . . , va′−2, va′ ). d Let v0, . . . , va′−2, wn : n<ω be distinct elements from X. For each n<ω let Vn ⊆ C be the ′ set of all v such that qa′ (v0, . . . , va′−2, v)= ±qa′ (v0, . . . , va′−2, wn′ ) for all n

Theorem 3.3. Suppose κ ≤ 2ℵ0 is a regular cardinal, and X ⊆ Rd has |X| = κ. Then there is Y ⊆ X of size κ, and some 2 ≤ a ≤ d + 1, such that Y is strictly a-rainbow.

9 Proof. By Theorem 2.4 we can choose Y ⊆ X of size κ such that Y is ∆d-indiscernible; then we conclude by Lemma 3.2.

For singular cardinals we do not have such an explicit conclusion, although we can say something about what the configurations look like:

Theorem 3.4. Suppose X ⊆ Rd, and E is an equivalence relation on X with infinitely many d classes, each class infinite. Suppose that considered as a subset of C , we have that X is (∆d, E)- indiscernible. Then there is 2 ≤ a∗ ≤ d+1 such that each E-equivalence class is strictly a∗-rainbow, and X is strongly a-rainbow for all a ≤ a∗.

Proof. There is some 2 ≤ a∗d + 1 such that each E-equivalence class is strictly a∗-rainbow, by Lemma 3.2 and (∆d, E)-indiscernibility, so we just need to show that X is strongly a-rainbow for all a ≤ a∗. Suppose not; say a ≤ a∗ and u0, . . . , ua−1, v0, . . . , va−1 are a-tuples from X with the same volume (possibly 0). We can suppose ua−1 6= vi for any i < a. We claim we can arrange that ui = vi for all i < a − 1, and that ua−1Eva−1. Indeed, choose w0, . . . , wa−1 distinct elements of X such that wi = ui for all i < a − 1, and wa−1 is some new element with wa−1Eua−1. By (∆, E)-indiscernibility, v has the same volume as both u and w, so the latter two have the same volume. So replace v by w. Let I ⊆ a be the set of all i < a with uiEua. Then clearly for any w0 . . . wa−1, if wi = ui for all i < a with i 6∈ I, and if wiEui for all i ∈ I, then u and w have the same volume. Moreover this ′ ′ ′ ′ holds whenever we replace u0, . . . , ua−1 by u0, . . . , ua−1, where uiEuj iff uiEuj. By reordering we can suppose I = {k, k + 1, . . . , a − 1}, for some k < a. Now k > 0, since otherwise u, v are a subset of a single equivalence class, and so this contradicts the choice of a∗. For the contradiction, we suppose we have arranged to have {ui/E : i < a} of minimum size. ℓ ℓ ℓ∗ Choose elements ui : i < a, ℓ < ω as follows: having defined ui for each ℓ<ℓ∗, let ui : i < k be ℓ∗ ℓ∗ ℓ∗ some new elements such that ui Euj iff uiEuj, and for i < k, ui is not E-related to any previous ℓ uj or any uj. 2·d·(a−k) For each ℓ<ω let Aℓ ⊆ C be the pre-variety of all (vk, . . . , va−1, wk, . . . , wa−1) such that ′ ℓ′ ℓ′ ℓ′ ℓ′ d·(a−k) for all ℓ < ℓ, pa(u0 , . . . , uk−1, vk, . . . , va−1, u0 , . . . , uk−1, wk, . . . , wa−1) = 0 (for tuples in R ℓ′ ℓ′ recall this is equivalent to saying that u0 , . . . , uk−1, vk, . . . , va−1 and ℓ′ ℓ′ u0 , . . . , uk−1, wk, . . . , wa−1 have the same volume). This is a descending chain of pre-varieties; but it must also be strict: for let ℓ<ω, and let ℓ vk, . . . , va−1, wk, . . . , wa−1 be new elements with each vi, wjEu0. Since {ui/E : i < a} was chosen of minimal size we must have that (vk, . . . , va−1, wk, . . . , wa−1) ∈ Aℓ\Aℓ+1. But this contradicts Hilbert’s Basis theorem.

Note that as a special case we have recovered Erd˝os’s Theorem 1.1: whenever X ⊆ Rd is infinite, there is a 2-rainbow Y ⊆ X with |Y | = |X|. This is because we must have a∗ ≥ 2.

4 Perfect subsets of Rd have rainbow perfect subsets

This section is in ZF+DC.

10 In this section we show that if X ⊆ Rd is perfect, then there is some perfect Y ⊆ X which is strictly a-rainbow for some a ≤ d + 1. Since Y is perfect we get |Y | = |X|.

Definition 4.1. We make several definitions.

1. Suppose P is a . A coloring f : P → [c] has the Baire property if, for all i ∈ [i], f −1(i) has the Baire property.

2. If x,y ∈ 2ω then ∆(x,y) = min{i : x(i) 6= y(i)}.

2ω 3. if u ∈ a , writing u = {x1,...,xa} in lexicographically increasing order (as always), then  2ω say that u is skew if for all 1 ≤ i < j < a, ∆(xi,xi+1) 6= ∆(xj,xj+1). Let be the set ω a skew of all u ∈ 2 which are skew. a  ω 4. We define f : 2 → LO([a−1]) as follows, where LO([a−1]) is the set of linear orders ∗ a skew of a − 1 (of which there are (a − 1)!). Namely let f∗(x0,...,xa−1) be the linear ordering < of a − 1 given by: i < j iff ∆(xi,xi+1) < ∆(xj,xj+1). Here we are writing u = {x0,...,xa−1} in increasing lexicographic order.

5. A perfect subtree of 2<ω is a subtree T of 2<ω (nonempty and closed under initial segments) such that for every s ∈ T there are t0,t1 ∈ T with s ⊂ t0,t1 and such that t0 and t1 are incompatible. Note that the set of branches [T ] through T is a perfect subset of 2ω, and this characterizes the perfect subsets of 2ω. Perfect subsets of 2ω are also called Cantor sets. We say that a Cantor set C is skew if for all x 6= y, x′ 6= y′ elements from C, if ∆(x,y)=∆(x′,y′) ω then {x,y} = {x′,y′}. In particular C ⊆ 2 for each a. a  a skew The following is due to Blass [1].

2ω Theorem 4.2. Suppose a, c are natural numbers and f : a → [c] has the Baire property. Then ω C there exists a skew Cantor set C ⊆ 2 so that for all u, v ∈ a , if f∗(u)= f∗(v) then f(u)= f(v).  (a−1)! Thus f ↾ C takes on only (a − 1)! values, and in fact there are only c possibilities for f ↾ C . ( a ) ( a )

A set X ⊆ Rd has the property if X is either countable or else has a perfect subset. This is a regularity property of subsets of Rd, and so holds for all reasonably definable subsets. For instance, every has the perfect set property: see Theorem 12.2 of [7]. Also, assuming sufficient large cardinals (for instance, infinitely many Woodin cardinals with a measurable cardinal above), all projective subsets of Rd have this property; see Theorem 32.14 of [7]. Moreover, if we let P SP denote the assertion that every subset of Rd has the perfect set property, then ZF + DC + P SP is consistent relative to an inaccessible cardinal (this is part of Solovay’s theorem, see Theorem 11.11 of [7]).

Theorem 4.3. Suppose P ⊆ Rd is perfect. Then there is a perfect set Q ⊆ X and some 2 ≤ a ≤ d + 1 such that Q is strictly a-rainbow.

Proof. It is not hard to find a continuous injection ρ : 2ω → P , such that the image of ρ is closed. Note that if C ⊆ 2ω is a Cantor set then ρ[C] is perfect.

11 ω 2 ′ Define f : 2a → [c] (for c large) so that f(u) codes the following information: for each a ≤ a,  2a and for each I, J ∈ ′ , whether or not the volume of ρ[uI ] is equal to 0, and whether or not the a  volume of ρ[uI ] is equal to the volume of ρ[uJ ]. Clearly we can choose f to have the Baire property (in fact, its graph will be Borel). Let C ⊆ 2ω be a skew Cantor set as in Theorem 4.2. It suffices to show that there some a ≤ d+1 such that ρ[C] is strictly a-rainbow. Suppose for some a ≤ d + 1 and for some x0 ∆(xi,xi+1) for all i < a − 1. Then whenever x ↾n= ρ(x0) ↾n, we get that the volume of (ρ(x), ρ(x1), . . . , ρ(xa−1)) is equal to 0. Hence {ρ(x) : x ∈ C,x ↾n= x0 ↾n} is contained in an a − 2-dimensional hyperplane; hence whenever C u = {yi : i ≤ a}∈ is such that each yi ↾n= ρ(x ) ↾n, we get that ρ[u] has volume 0. Since every a  0 element of C has the same type as some such u, we get that for all u ∈ C , ρ[u] has volume 0, a  a  and so ρ[C] is contained in an a − 2-dimensional hyperplane. Let a be largest so that this fails; thus ρ[C] is a subset of an a − 1-dimensional hyperplane, but for each a′ ≤ a and for each u ∈ C , ρ[u] has nonzero volume. We claim that for each a′ ≤ a, ρ[C] a  is a′-rainbow (and hence strongly a’-rainbow). ′ Suppose not, say a ≤ a and x0 ∆(xi,xj) and N > ∆(yi,yj) for all i < j < a , and whenever ′ ′ xi 6= yj then N > ∆(xi,yj). Now choose i∗ < a such that xi∗ 6∈ {yi : i < a }; then for any x with ′ ′ x ↾N = xi∗ ↾N we have that the a -ary volume of ρ[x0,...,xi∗−1,xi∗ ,xi∗+1,...,xa −1] is equal to the ′ ′ ′ a − 1-ary volume of ρ[x0,...,xi∗−1,x,xi∗+1,...,xa −1], both being equal to the a − 1-ary volume of y0,...,ya′−1. ′ Given z with f∗(z)= f∗(x), let Nz be the maximum of ∆(zi, zj)+1: i < j < a . Let the cone ′ above z, Cz, be all z such that z ↾Nz = zi∗ ↾Nz . Then for any z ∈Cz, ρ[z] has the same a -ary volume as ρ[z\{zi∗ } ∪ {z}]. Recall that qa′ (v0, . . . , va′−1, w0, . . . , wa′−1) is a polynomial (each vi, wj is a tuple of d-variables) d ′ such that given α0,...,αa′−1, β0,...,βa′−1 from R , α and β have the same a − 1-ary volume iff d ′ qa′ (α, β) = 0; given tuples α, β from C , then we define them to have the same a − 1-ary volume if qa′ (α, β) = 0. n n n n n Inductively choose x : n<ω so that each x = x0

Corollary 4.4.

1. Suppose X ⊆ Rd is analytic and uncountable. Then there is a perfect set Q ⊆ X and some 2 ≤ a ≤ d + 1 such that Q is strictly a-rainbow.

12 2. Assume sufficient large cardinals. Suppose X ⊆ Rd is projective and uncountable. Then there is a perfect set Q ⊆ X and some 2 ≤ a ≤ d + 1 such that Q is strictly a-rainbow.

3. Assume PSP. Suppose X ⊆ Rd is uncountable. Then there is a perfect set Q ⊆ X and some 2 ≤ a ≤ d + 1 such that Q is strictly a-rainbow.

5 A model of set theory where an of reals has no uncountable 2-rainbow subset

In this section we prove it is consistent with ZF that there is an uncountable set of reals without an uncountable 2-rainbow subset. Thus some amount of choice is necessary. The proof is a standard symmetric models argument; for a source on this, see [6], Chapter 15. We attempted to prove consistency over ZF + DC, but could not, so we leave the following as an open question: Question. Is it consistent with ZF + DC that there is an uncountable set of reals without an uncountable 2-rainbow subset?

Theorem 5.1. Suppose V |= ZFC. Then there is a symmetric submodel M of a forcing extension V[G] of V, such that M |= ZF + there is an uncountable set of reals with no uncountable 2-rainbow subset.

Proof. We identify x ∈ 2ω with the element of [0, 1] with binary expansion given by x. (The collisions do not matter.) Let P be the forcing notion of all finite partial functions from ω × ω → 2. Then forcing by P adds a Cohen-generic f˙ ∈ (2ω×ω). For each n<ω, leta ˙ n be a P -name for {m<ω : f˙(n,m) = 1}, so each 0P a˙ n ⊆ ω. For each s ⊂ ω finite leta ˙ n,s be a P -name fora ˙ n∆s (the symmetric difference). Let A˙ be a P -name for {a˙ n,s : n<ω,s ⊂ ω finite}. Let G be the group of all permutations σ of ω × ω × 2 such that: there is a permutation σ0 of ω, such that for all (n,m,i), σ(n,m,i) = (σ0(n),m,j) for some j (thus we have a map σ 7→ σ0). Each σ ∈ G induces an automorphism of P and hence of its Boolean completion B, which we identify with σ. Let F be the filter of subgroups on G, generated by Fix(˙an) for each n<ω, along with ˙ ˙ Fix(A). Note that σ ∈ Fix(˙an) iff σ ↾{n}×ω×2 is the identity, and σ ∈ Fix(A) iff for each n<ω, there are only finitely many m with σ(n,m, 0) 6= (σ0(n), m, 0). Let V[G] be a forcing extension by P , and let M be the symmetric submodel determined by F, G, that is M is the set of alla ˙ V[G], fora ˙ a P -name such that Fix(˙a) ∈ F (as computed in V), and moreover this holds hereditarily. M is a model of ZF ; see Chapter 15 of [6]. Now A˙ V[G] ∈ M since Fix(A˙) ∈ F by construction and this also holds for eacha ˙ ∈ A˙. We claim that M, A˙V[G] works. It is well known that A˙ V[G] is uncountable in V[G]: if we look at the larger model N := V[G] V({a˙ n : n ∈ ω) then this is the standard example, due to Cohen, of a model of ZF where choice V[G] fails; {a˙ n : n ∈ ω} is an uncountable set and in fact it has no countable subset. See Chapter 14 of [6]. So the larger set A˙ must still be uncountable in the smaller model M.

13 Suppose towards a contradiction that in M, A˙ V[G] had an uncountable 2-rainbow subset. Then we can choose some hereditarily F-symmetric P -name B˙ and some p ∈ P such that p forces: B˙ is an uncountable 2-rainbow subset of A˙. We can choose N large enough so that dom(p) ⊂ N × ω and, for every σ ∈ G, if σ ↾N×ω×2 is the identity and if σ ∈ Fix(A˙), then σ(B˙ )= B˙ . For each n<ω and s ⊂ ω finite let Qn be the set of all q ≤ p such that q forces: B˙ ∩ {a˙ n,s : s ⊂ ω finite}= 6 ∅. We claim that each Qn is nonempty, i.e. p does not force that B˙ ∩ {a˙ n,s : s ⊂ ω finite} is empty. ′ For suppose it did; then for every n ≥ N, p forces that B˙ ∩ {a˙ n′,s : s ⊂ ω finite} is empty (by considering σ ∈ G that fix the second and third coordinates and interchange n,n′). But then p would force that B˙ ⊆ {a˙ n,s : n

6 Acknowledgements

We want to thank Carolyn Gasarch whose skepticism of the was one of our inspi- rations. We also want to thank our co-authors on the original Distinct Volume Sets paper, (David Conlon, Jacob Fox, David Harris, and Sam Zbarsky) since that paper was also one of our inspi- rations. Finally, we would like to thank Chris Laskowski for greatly improving the exposition of Theorem 2.5.

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